WB JEE 2026 | Trigonometric Functions & Equations Question 30 | Mathematics

WB JEE 2026

MCQ (Single Correct Answer)

-0.25

$\theta$ elimination from the equation $x^2+y^2=\frac{x \cos 3 \theta+y \sin 3 \theta}{\cos ^3 \theta}=\frac{y \cos 3 \theta-x \sin 3 \theta}{\sin ^3 \theta}$ will be

$4\left(x^4+y^4\right)=3 x+4 y$

$\left(x^2+y^2+2 x\right)\left(x^2+y^2-x\right)=2 y^2$

$\left(x^2+y^2-2 x\right)\left(x^2+y^2+x\right)=9 y$

$x^{2 / 3}+y^{2 / 3}=1$

WB JEE 2026

MCQ (Single Correct Answer)

-0.25

The general solution of the equation $\sin ^{100} \mathrm{x}-\cos ^{100} \mathrm{x}=1$ is

$\left\{2 n \pi+\frac{\pi}{3}: n \in I\right\}$

$\left\{n \pi+\frac{\pi}{4}: n \in I\right\}$

$\left\{n \pi \pm \frac{\pi}{2}: n \in I\right\}$

$\left\{2 \mathrm{n} \pi-\frac{\pi}{3}: \mathrm{n} \in \mathrm{I}\right\}$

WB JEE 2025

MCQ (Single Correct Answer)

-0.25

Let $f_n(x)=\tan \frac{x}{2}(1+\sec x)(1+\sec 2 x) \ldots\left(1+\sec 2^n x\right)$, then

$f_5\left(\frac{\pi}{16}\right)=1$

$f_4\left(\frac{\pi}{16}\right)=1$

$f_3\left(\frac{\pi}{16}\right)=1$

$f_2\left(\frac{\pi}{16}\right)=1$

WB JEE 2025

MCQ (Single Correct Answer)

-0.5

If $\cos (\theta+\phi)=\frac{3}{5}$ and $\sin (\theta-\phi)=\frac{5}{13}, 0<\theta, \phi<\frac{\pi}{4}$, then $\cot (2 \theta)$ has the value

$\frac{16}{63}$

$\frac{63}{16}$

$\frac{3}{13}$

$\frac{13}{3}$

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